# Quantum Bayesian Networks

## October 1, 2014

### Monty Hall Problem, as discussed by a 5 year old Illuminati

Filed under: Uncategorized — rrtucci @ 9:52 pm

A friend just sent me a link to the following NYT article:

The Odds, Continually Updated by Flim D. Flam (would I lie to you?) Sept. 29, 2014, New York Times

Even though the article doesn’t mention the obvious fact that quantum computers will someday soon revolutionize the Bayesian field (because they will be able to do Bayesian calculations much faster than classical computers), it’s not such a hopelessly outdated, clueless piece of reporting. Really. The article doth sing the praises of Bayesian techniques, so it ain’t that utterly bad.

The article makes a big deal about the Monty Hall problem. It’s a cute problem, for a five years old. Correction: A five year old that has been taught Bayesian networks. We B net advocates teach our children Monty Hall on day one, of pre-kindergarten. It’s quite conceivable that an intelligent 50 year old kid who has been taught only the frequentist canon might spend hours on such a problem, and ultimately give up.

Here is how a 5 year old illuminati who is conversant with the language of B nets would discuss such a piddling problem:

Let $\theta({\cal S})$ stand for the “truth function”. It equals 1 if statement ${\cal S}$ is true and 0 otherwise. For example, $\theta(a=b)=\delta_a^b$ is the Kronecker delta function.

The Monty Hall problem can be modelled by the B net shown, where

$c=$ the door behind which the car actually is.
$y=$ the door opened by you (the contestant), on your first selection.
$m=$ the door opened by Monty (game host)

If we label the doors 1,2,3, then $m,c,y\in \{1,2,3\}$ and

$P(c)=\frac{1}{3}$

$P(y)=\frac{1}{3}$

$P(m|c,y)=\frac{1}{2}\theta(m\neq c)\theta(y=c)\;\;\;+\;\;\;\theta(m\neq y,c)\theta(y\neq c)$

It’s easy to show that the above node probabilities imply that

$P(c=1|m=2,y=1)=\frac{1}{3}$

$P(c=3|m=2,y=1)=\frac{2}{3}$

So you are twice as likely to win if you switch your final selection to be the door which is neither your first choice nor Monty’s choice.